MC
6cd3_8c56
A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.
| Gene F is related to the 'fuzzy' phenotype. A fruit fly that is homozygous recessive for Gene F is covered in a dense layer of hairs, giving it a soft appearance. | |
| Gene Y is linked with the 'yucky' phenotype. A fruit fly that is homozygous recessive for Gene Y gives off an unpleasant odor and has a generally unappealing look. |
| Phenotype | Genotypes | Progeny Count | |
|---|---|---|---|
| fuzzy, yucky | f | y | 41,500 |
| fuzzy | f | + | 83,205 |
| yucky | + | y | 83,370 |
| wildtype | + | + | 41,925 |
| TOTAL = | 250,000 | ||
The resulting phenotypes are summarized in the table above.
With the progeny data from the table, calculate the genetic distance between the two genes, expressing your answer in centimorgans (cM)
41,500/250,000 = 41,500/250,000 = 0.1660 = 16.60 cM Incorrect 41,925/250,000 = 41,925/250,000 = 0.1677 = 16.77 cM Incorrect 83,205/250,000 = 83,205/250,000 = 0.3328 = 33.28 cM Incorrect 83,370/250,000 = 83,370/250,000 = 0.3335 = 33.35 cM Incorrect (41,500 + 41,925)/250,000 = 83,425/250,000 = 0.3337 = 33.37 cM Correct (41,500 + 83,205)/250,000 = 124,705/250,000 = 0.4988 = 49.88 cM Incorrect MC96fc_75f2
A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.
| Gene H is associated with the 'horsey' phenotype. A fruit fly that is homozygous recessive for Gene H is quite big and strong-looking, much larger than your typical fruit fly. | |
| Gene K is analogous to the 'kidney' phenotype. A fruit fly that is homozygous recessive for Gene K has a body shape that is curved, similar to a kidney bean. |
| Phenotype | Genotypes | Progeny Count | |
|---|---|---|---|
| horsey, kidney | h | k | 35,621 |
| horsey | h | + | 14,556 |
| kidney | + | k | 14,214 |
| wildtype | + | + | 35,609 |
| TOTAL = | 100,000 | ||
The resulting phenotypes are summarized in the table above.
With the progeny data from the table, calculate the genetic distance between the two genes, expressing your answer in centimorgans (cM)
14,214/100,000 = 14,214/100,000 = 0.1421 = 14.21 cM Incorrect 14,556/100,000 = 14,556/100,000 = 0.1456 = 14.56 cM Incorrect (14,214 + 14,556)/100,000 = 28,770/100,000 = 0.2877 = 28.77 cM Correct 35,609/100,000 = 35,609/100,000 = 0.3561 = 35.61 cM Incorrect 35,621/100,000 = 35,621/100,000 = 0.3562 = 35.62 cM Incorrect (14,214 + 35,609)/100,000 = 49,823/100,000 = 0.4982 = 49.82 cM Incorrect MC977b_bb0e
A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.
| Gene F is related to the 'fuzzy' phenotype. A fruit fly that is homozygous recessive for Gene F is covered in a dense layer of hairs, giving it a soft appearance. | |
| Gene Y is analogous to the 'yucky' phenotype. A fruit fly that is homozygous recessive for Gene Y gives off an unpleasant odor and has a generally unappealing look. |
| Phenotype | Genotypes | Progeny Count | |
|---|---|---|---|
| fuzzy, yucky | f | y | 16,720 |
| fuzzy | f | + | 5,632 |
| yucky | + | y | 5,690 |
| wildtype | + | + | 16,958 |
| TOTAL = | 45,000 | ||
The resulting phenotypes are summarized in the table above.
With the progeny data from the table, calculate the genetic distance between the two genes, expressing your answer in centimorgans (cM)
5,632/45,000 = 5,632/45,000 = 0.1252 = 12.52 cM Incorrect 5,690/45,000 = 5,690/45,000 = 0.1264 = 12.64 cM Incorrect (5,632 + 5,690)/45,000 = 11,322/45,000 = 0.2516 = 25.16 cM Correct 16,720/45,000 = 16,720/45,000 = 0.3716 = 37.16 cM Incorrect 16,958/45,000 = 16,958/45,000 = 0.3768 = 37.68 cM Incorrect (5,632 + 16,958)/45,000 = 22,590/45,000 = 0.5020 = 50.20 cM Incorrect MC6de2_4e9b
A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.
| Gene K is correlated with the 'kidney' phenotype. A fruit fly that is homozygous recessive for Gene K has a body shape that is curved, similar to a kidney bean. | |
| Gene N is connected with the 'nerdy' phenotype. A fruit fly that is homozygous recessive for Gene N has large, prominent eyes that stand out, much like thick-rimmed glasses. |
| Phenotype | Genotypes | Progeny Count | |
|---|---|---|---|
| kidney, nerdy | k | n | 36,640 |
| kidney | k | + | 108,632 |
| nerdy | + | n | 108,085 |
| wildtype | + | + | 36,643 |
| TOTAL = | 290,000 | ||
The resulting phenotypes are summarized in the table above.
With the progeny data from the table, calculate the genetic distance between the two genes, expressing your answer in centimorgans (cM)
36,640/290,000 = 36,640/290,000 = 0.1263 = 12.63 cM Incorrect 36,643/290,000 = 36,643/290,000 = 0.1264 = 12.64 cM Incorrect (36,640 + 36,643)/290,000 = 73,283/290,000 = 0.2527 = 25.27 cM Correct 108,085/290,000 = 108,085/290,000 = 0.3727 = 37.27 cM Incorrect 108,632/290,000 = 108,632/290,000 = 0.3746 = 37.46 cM Incorrect (36,640 + 108,632)/290,000 = 145,272/290,000 = 0.5009 = 50.09 cM Incorrect MCa3f5_0fe0
A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.
| Gene R is associated with the 'rusty' phenotype. A fruit fly that is homozygous recessive for Gene R has a reddish-brown color, much like rusted iron metal. | |
| Gene X is correlated with the 'xanthic' phenotype. A fruit fly that is homozygous recessive for Gene X has a fluorescent bright yellow coloring. |
| Phenotype | Genotypes | Progeny Count | |
|---|---|---|---|
| rusty, xanthic | r | x | 7,085 |
| rusty | r | + | 30,407 |
| xanthic | + | x | 30,508 |
| wildtype | + | + | 7,000 |
| TOTAL = | 75,000 | ||
The resulting phenotypes are summarized in the table above.
With the progeny data from the table, calculate the genetic distance between the two genes, expressing your answer in centimorgans (cM)
7,000/75,000 = 7,000/75,000 = 0.0933 = 9.33 cM Incorrect 7,085/75,000 = 7,085/75,000 = 0.0945 = 9.45 cM Incorrect (7,000 + 7,085)/75,000 = 14,085/75,000 = 0.1878 = 18.78 cM Correct 30,407/75,000 = 30,407/75,000 = 0.4054 = 40.54 cM Incorrect 30,508/75,000 = 30,508/75,000 = 0.4068 = 40.68 cM Incorrect (7,000 + 30,508)/75,000 = 37,508/75,000 = 0.5001 = 50.01 cM Incorrect MC913b_d0d5
A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.
| Gene H is linked with the 'horsey' phenotype. A fruit fly that is homozygous recessive for Gene H is quite big and strong-looking, much larger than your typical fruit fly. | |
| Gene M is linked with the 'mushy' phenotype. A fruit fly that is homozygous recessive for Gene M feels soft to the touch and unusually squishy, unlike the usual firmness. |
| Phenotype | Genotypes | Progeny Count | |
|---|---|---|---|
| horsey, mushy | h | m | 76,372 |
| horsey | h | + | 13,646 |
| mushy | + | m | 13,588 |
| wildtype | + | + | 76,394 |
| TOTAL = | 180,000 | ||
The resulting phenotypes are summarized in the table above.
With the progeny data from the table, calculate the genetic distance between the two genes, expressing your answer in centimorgans (cM)
13,588/180,000 = 13,588/180,000 = 0.0755 = 7.55 cM Incorrect 13,646/180,000 = 13,646/180,000 = 0.0758 = 7.58 cM Incorrect (13,588 + 13,646)/180,000 = 27,234/180,000 = 0.1513 = 15.13 cM Correct 76,372/180,000 = 76,372/180,000 = 0.4243 = 42.43 cM Incorrect 76,394/180,000 = 76,394/180,000 = 0.4244 = 42.44 cM Incorrect (13,588 + 76,394)/180,000 = 89,982/180,000 = 0.4999 = 49.99 cM Incorrect MC4dcf_cf5b
A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.
| Gene H is linked with the 'horsey' phenotype. A fruit fly that is homozygous recessive for Gene H is quite big and strong-looking, much larger than your typical fruit fly. | |
| Gene W is affiliated with the 'waxy' phenotype. A fruit fly that is homozygous recessive for Gene W has a thick protective layer that is water resistant and opague. |
| Phenotype | Genotypes | Progeny Count | |
|---|---|---|---|
| horsey, waxy | h | w | 2,187 |
| horsey | h | + | 7,807 |
| waxy | + | w | 7,783 |
| wildtype | + | + | 2,223 |
| TOTAL = | 20,000 | ||
The resulting phenotypes are summarized in the table above.
With the progeny data from the table, calculate the genetic distance between the two genes, expressing your answer in centimorgans (cM)
2,187/20,000 = 2,187/20,000 = 0.1094 = 10.94 cM Incorrect 2,223/20,000 = 2,223/20,000 = 0.1111 = 11.12 cM Incorrect (2,187 + 2,223)/20,000 = 4,410/20,000 = 0.2205 = 22.05 cM Correct 7,783/20,000 = 7,783/20,000 = 0.3891 = 38.91 cM Incorrect 7,807/20,000 = 7,807/20,000 = 0.3903 = 39.03 cM Incorrect (2,187 + 7,783)/20,000 = 9,970/20,000 = 0.4985 = 49.85 cM Incorrect MC85b0_a931
A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.
| Gene F is connected with the 'fuzzy' phenotype. A fruit fly that is homozygous recessive for Gene F is covered in a dense layer of hairs, giving it a soft appearance. | |
| Gene R is associated with the 'rusty' phenotype. A fruit fly that is homozygous recessive for Gene R has a reddish-brown color, much like rusted iron metal. |
| Phenotype | Genotypes | Progeny Count | |
|---|---|---|---|
| fuzzy, rusty | f | r | 73,692 |
| fuzzy | f | + | 31,161 |
| rusty | + | r | 31,062 |
| wildtype | + | + | 74,085 |
| TOTAL = | 210,000 | ||
The resulting phenotypes are summarized in the table above.
With the progeny data from the table, calculate the genetic distance between the two genes, expressing your answer in centimorgans (cM)
31,062/210,000 = 31,062/210,000 = 0.1479 = 14.79 cM Incorrect 31,161/210,000 = 31,161/210,000 = 0.1484 = 14.84 cM Incorrect (31,062 + 31,161)/210,000 = 62,223/210,000 = 0.2963 = 29.63 cM Correct 73,692/210,000 = 73,692/210,000 = 0.3509 = 35.09 cM Incorrect 74,085/210,000 = 74,085/210,000 = 0.3528 = 35.28 cM Incorrect (31,161 + 73,692)/210,000 = 104,853/210,000 = 0.4993 = 49.93 cM Incorrect MC62ea_b331
A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.
| Gene C is associated with the 'chummy' phenotype. A fruit fly that is homozygous recessive for Gene C shows behavior where it always maintains a close distance to other flies. | |
| Gene J is correlated with the 'jerky' phenotype. A fruit fly that is homozygous recessive for Gene J moves in rapid and sudden movements, displaying an unpredictable flight pattern. |
| Phenotype | Genotypes | Progeny Count | |
|---|---|---|---|
| chummy, jerky | c | j | 1,297 |
| chummy | c | + | 12,691 |
| jerky | + | j | 12,761 |
| wildtype | + | + | 1,251 |
| TOTAL = | 28,000 | ||
The resulting phenotypes are summarized in the table above.
With the progeny data from the table, calculate the genetic distance between the two genes, expressing your answer in centimorgans (cM)
1,251/28,000 = 1,251/28,000 = 0.0447 = 4.47 cM Incorrect 1,297/28,000 = 1,297/28,000 = 0.0463 = 4.63 cM Incorrect (1,251 + 1,297)/28,000 = 2,548/28,000 = 0.0910 = 9.10 cM Correct 12,691/28,000 = 12,691/28,000 = 0.4532 = 45.32 cM Incorrect 12,761/28,000 = 12,761/28,000 = 0.4557 = 45.57 cM Incorrect (1,297 + 12,761)/28,000 = 14,058/28,000 = 0.5021 = 50.21 cM Incorrect MC79d1_7f8e
A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.
| Gene C is linked with the 'chummy' phenotype. A fruit fly that is homozygous recessive for Gene C shows behavior where it always maintains a close distance to other flies. | |
| Gene T is linked with the 'tipsy' phenotype. A fruit fly that is homozygous recessive for Gene T moves in an erratic path, suggesting a lack of coordination, as if intoxicated. |
| Phenotype | Genotypes | Progeny Count | |
|---|---|---|---|
| chummy, tipsy | c | t | 25,829 |
| chummy | c | + | 14,163 |
| tipsy | + | t | 14,149 |
| wildtype | + | + | 25,859 |
| TOTAL = | 80,000 | ||
The resulting phenotypes are summarized in the table above.
With the progeny data from the table, calculate the genetic distance between the two genes, expressing your answer in centimorgans (cM)
14,149/80,000 = 14,149/80,000 = 0.1769 = 17.69 cM Incorrect 14,163/80,000 = 14,163/80,000 = 0.1770 = 17.70 cM Incorrect 25,829/80,000 = 25,829/80,000 = 0.3229 = 32.29 cM Incorrect 25,859/80,000 = 25,859/80,000 = 0.3232 = 32.32 cM Incorrect (14,149 + 14,163)/80,000 = 28,312/80,000 = 0.3539 = 35.39 cM Correct (14,163 + 25,829)/80,000 = 39,992/80,000 = 0.4999 = 49.99 cM Incorrect MC81c3_85b5
A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.
| Gene K is linked with the 'kidney' phenotype. A fruit fly that is homozygous recessive for Gene K has a body shape that is curved, similar to a kidney bean. | |
| Gene M is related to the 'mushy' phenotype. A fruit fly that is homozygous recessive for Gene M feels soft to the touch and unusually squishy, unlike the usual firmness. |
| Phenotype | Genotypes | Progeny Count | |
|---|---|---|---|
| kidney, mushy | k | m | 20,614 |
| kidney | k | + | 4,289 |
| mushy | + | m | 4,441 |
| wildtype | + | + | 20,656 |
| TOTAL = | 50,000 | ||
The resulting phenotypes are summarized in the table above.
With the progeny data from the table, calculate the genetic distance between the two genes, expressing your answer in centimorgans (cM)
4,289/50,000 = 4,289/50,000 = 0.0858 = 8.58 cM Incorrect 4,441/50,000 = 4,441/50,000 = 0.0888 = 8.88 cM Incorrect (4,289 + 4,441)/50,000 = 8,730/50,000 = 0.1746 = 17.46 cM Correct 20,614/50,000 = 20,614/50,000 = 0.4123 = 41.23 cM Incorrect 20,656/50,000 = 20,656/50,000 = 0.4131 = 41.31 cM Incorrect (4,289 + 20,614)/50,000 = 24,903/50,000 = 0.4981 = 49.81 cM Incorrect MC3549_861e
A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.
| Gene A is linked with the 'artsy' phenotype. A fruit fly that is homozygous recessive for Gene A has wings that are colorful and distinctive patterns. | |
| Gene M is correlated with the 'mushy' phenotype. A fruit fly that is homozygous recessive for Gene M feels soft to the touch and unusually squishy, unlike the usual firmness. |
| Phenotype | Genotypes | Progeny Count | |
|---|---|---|---|
| artsy, mushy | a | m | 11,925 |
| artsy | a | + | 23,140 |
| mushy | + | m | 23,291 |
| wildtype | + | + | 11,644 |
| TOTAL = | 70,000 | ||
The resulting phenotypes are summarized in the table above.
With the progeny data from the table, calculate the genetic distance between the two genes, expressing your answer in centimorgans (cM)
11,644/70,000 = 11,644/70,000 = 0.1663 = 16.63 cM Incorrect 11,925/70,000 = 11,925/70,000 = 0.1704 = 17.04 cM Incorrect 23,140/70,000 = 23,140/70,000 = 0.3306 = 33.06 cM Incorrect 23,291/70,000 = 23,291/70,000 = 0.3327 = 33.27 cM Incorrect (11,644 + 11,925)/70,000 = 23,569/70,000 = 0.3367 = 33.67 cM Correct (11,925 + 23,140)/70,000 = 35,065/70,000 = 0.5009 = 50.09 cM Incorrect MC262a_c931
A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.
| Gene P is related to the 'prickly' phenotype. A fruit fly that is homozygous recessive for Gene P is covered with sharp bristles, giving it a spiky texture. | |
| Gene W is related to the 'waxy' phenotype. A fruit fly that is homozygous recessive for Gene W has a thick protective layer that is water resistant and opague. |
| Phenotype | Genotypes | Progeny Count | |
|---|---|---|---|
| prickly, waxy | p | w | 83,786 |
| prickly | p | + | 21,178 |
| waxy | + | w | 21,389 |
| wildtype | + | + | 83,647 |
| TOTAL = | 210,000 | ||
The resulting phenotypes are summarized in the table above.
With the progeny data from the table, calculate the genetic distance between the two genes, expressing your answer in centimorgans (cM)
21,178/210,000 = 21,178/210,000 = 0.1008 = 10.08 cM Incorrect 21,389/210,000 = 21,389/210,000 = 0.1019 = 10.19 cM Incorrect (21,178 + 21,389)/210,000 = 42,567/210,000 = 0.2027 = 20.27 cM Correct 83,647/210,000 = 83,647/210,000 = 0.3983 = 39.83 cM Incorrect 83,786/210,000 = 83,786/210,000 = 0.3990 = 39.90 cM Incorrect (21,389 + 83,786)/210,000 = 105,175/210,000 = 0.5008 = 50.08 cM Incorrect MC4c85_8113
A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.
| Gene B is linked with the 'bumpy' phenotype. A fruit fly that is homozygous recessive for Gene B has a skin texture that is not smooth, but rough with small bumps all over. | |
| Gene K is connected with the 'kidney' phenotype. A fruit fly that is homozygous recessive for Gene K has a body shape that is curved, similar to a kidney bean. |
| Phenotype | Genotypes | Progeny Count | |
|---|---|---|---|
| bumpy, kidney | b | k | 87,343 |
| bumpy | b | + | 17,700 |
| kidney | + | k | 17,349 |
| wildtype | + | + | 87,608 |
| TOTAL = | 210,000 | ||
The resulting phenotypes are summarized in the table above.
With the progeny data from the table, calculate the genetic distance between the two genes, expressing your answer in centimorgans (cM)
17,349/210,000 = 17,349/210,000 = 0.0826 = 8.26 cM Incorrect 17,700/210,000 = 17,700/210,000 = 0.0843 = 8.43 cM Incorrect (17,349 + 17,700)/210,000 = 35,049/210,000 = 0.1669 = 16.69 cM Correct 87,343/210,000 = 87,343/210,000 = 0.4159 = 41.59 cM Incorrect 87,608/210,000 = 87,608/210,000 = 0.4172 = 41.72 cM Incorrect (17,349 + 87,343)/210,000 = 104,692/210,000 = 0.4985 = 49.85 cM Incorrect MC780c_88bd
A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.
| Gene M is analogous to the 'mushy' phenotype. A fruit fly that is homozygous recessive for Gene M feels soft to the touch and unusually squishy, unlike the usual firmness. | |
| Gene N is associated with the 'nerdy' phenotype. A fruit fly that is homozygous recessive for Gene N has large, prominent eyes that stand out, much like thick-rimmed glasses. |
| Phenotype | Genotypes | Progeny Count | |
|---|---|---|---|
| mushy, nerdy | m | n | 45,711 |
| mushy | m | + | 14,175 |
| nerdy | + | n | 14,457 |
| wildtype | + | + | 45,657 |
| TOTAL = | 120,000 | ||
The resulting phenotypes are summarized in the table above.
With the progeny data from the table, calculate the genetic distance between the two genes, expressing your answer in centimorgans (cM)
14,175/120,000 = 14,175/120,000 = 0.1181 = 11.81 cM Incorrect 14,457/120,000 = 14,457/120,000 = 0.1205 = 12.05 cM Incorrect (14,175 + 14,457)/120,000 = 28,632/120,000 = 0.2386 = 23.86 cM Correct 45,657/120,000 = 45,657/120,000 = 0.3805 = 38.05 cM Incorrect 45,711/120,000 = 45,711/120,000 = 0.3809 = 38.09 cM Incorrect (14,457 + 45,711)/120,000 = 60,168/120,000 = 0.5014 = 50.14 cM Incorrect MC60b7_5042
A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.
| Gene F is linked with the 'fuzzy' phenotype. A fruit fly that is homozygous recessive for Gene F is covered in a dense layer of hairs, giving it a soft appearance. | |
| Gene R is correlated with the 'rusty' phenotype. A fruit fly that is homozygous recessive for Gene R has a reddish-brown color, much like rusted iron metal. |
| Phenotype | Genotypes | Progeny Count | |
|---|---|---|---|
| fuzzy, rusty | f | r | 35,129 |
| fuzzy | f | + | 95,000 |
| rusty | + | r | 94,462 |
| wildtype | + | + | 35,409 |
| TOTAL = | 260,000 | ||
The resulting phenotypes are summarized in the table above.
With the progeny data from the table, calculate the genetic distance between the two genes, expressing your answer in centimorgans (cM)
35,129/260,000 = 35,129/260,000 = 0.1351 = 13.51 cM Incorrect 35,409/260,000 = 35,409/260,000 = 0.1362 = 13.62 cM Incorrect (35,129 + 35,409)/260,000 = 70,538/260,000 = 0.2713 = 27.13 cM Correct 94,462/260,000 = 94,462/260,000 = 0.3633 = 36.33 cM Incorrect 95,000/260,000 = 95,000/260,000 = 0.3654 = 36.54 cM Incorrect (35,129 + 94,462)/260,000 = 129,591/260,000 = 0.4984 = 49.84 cM Incorrect MC84b7_23b2
A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.
| Gene C is analogous to the 'chummy' phenotype. A fruit fly that is homozygous recessive for Gene C shows behavior where it always maintains a close distance to other flies. | |
| Gene E is correlated with the 'eery' phenotype. A fruit fly that is homozygous recessive for Gene E appears to have something off, crooked limbs and other twisted appendages. |
| Phenotype | Genotypes | Progeny Count | |
|---|---|---|---|
| chummy, eery | c | e | 4,515 |
| chummy | c | + | 10,548 |
| eery | + | e | 10,536 |
| wildtype | + | + | 4,401 |
| TOTAL = | 30,000 | ||
The resulting phenotypes are summarized in the table above.
With the progeny data from the table, calculate the genetic distance between the two genes, expressing your answer in centimorgans (cM)
4,401/30,000 = 4,401/30,000 = 0.1467 = 14.67 cM Incorrect 4,515/30,000 = 4,515/30,000 = 0.1505 = 15.05 cM Incorrect (4,401 + 4,515)/30,000 = 8,916/30,000 = 0.2972 = 29.72 cM Correct 10,536/30,000 = 10,536/30,000 = 0.3512 = 35.12 cM Incorrect 10,548/30,000 = 10,548/30,000 = 0.3516 = 35.16 cM Incorrect (4,401 + 10,548)/30,000 = 14,949/30,000 = 0.4983 = 49.83 cM Incorrect MC2f6d_2f0f
A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.
| Gene A is analogous to the 'artsy' phenotype. A fruit fly that is homozygous recessive for Gene A has wings that are colorful and distinctive patterns. | |
| Gene B is affiliated with the 'bumpy' phenotype. A fruit fly that is homozygous recessive for Gene B has a skin texture that is not smooth, but rough with small bumps all over. |
| Phenotype | Genotypes | Progeny Count | |
|---|---|---|---|
| artsy, bumpy | a | b | 618 |
| artsy | a | + | 314 |
| bumpy | + | b | 307 |
| wildtype | + | + | 561 |
| TOTAL = | 1,800 | ||
The resulting phenotypes are summarized in the table above.
With the progeny data from the table, calculate the genetic distance between the two genes, expressing your answer in centimorgans (cM)
307/1,800 = 307/1,800 = 0.1706 = 17.06 cM Incorrect 314/1,800 = 314/1,800 = 0.1744 = 17.44 cM Incorrect 561/1,800 = 561/1,800 = 0.3117 = 31.17 cM Incorrect 618/1,800 = 618/1,800 = 0.3433 = 34.33 cM Incorrect (307 + 314)/1,800 = 621/1,800 = 0.3450 = 34.50 cM Correct (314 + 618)/1,800 = 932/1,800 = 0.5178 = 51.78 cM Incorrect MC502c_8542
A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.
| Gene C is connected with the 'chummy' phenotype. A fruit fly that is homozygous recessive for Gene C shows behavior where it always maintains a close distance to other flies. | |
| Gene T is affiliated with the 'tipsy' phenotype. A fruit fly that is homozygous recessive for Gene T moves in an erratic path, suggesting a lack of coordination, as if intoxicated. |
| Phenotype | Genotypes | Progeny Count | |
|---|---|---|---|
| chummy, tipsy | c | t | 47,052 |
| chummy | c | + | 8,283 |
| tipsy | + | t | 8,184 |
| wildtype | + | + | 46,481 |
| TOTAL = | 110,000 | ||
The resulting phenotypes are summarized in the table above.
With the progeny data from the table, calculate the genetic distance between the two genes, expressing your answer in centimorgans (cM)
8,184/110,000 = 8,184/110,000 = 0.0744 = 7.44 cM Incorrect 8,283/110,000 = 8,283/110,000 = 0.0753 = 7.53 cM Incorrect (8,184 + 8,283)/110,000 = 16,467/110,000 = 0.1497 = 14.97 cM Correct 46,481/110,000 = 46,481/110,000 = 0.4226 = 42.26 cM Incorrect 47,052/110,000 = 47,052/110,000 = 0.4277 = 42.77 cM Incorrect (8,184 + 47,052)/110,000 = 55,236/110,000 = 0.5021 = 50.21 cM Incorrect MC3f39_5189
A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.
| Gene B is affiliated with the 'bumpy' phenotype. A fruit fly that is homozygous recessive for Gene B has a skin texture that is not smooth, but rough with small bumps all over. | |
| Gene F is linked with the 'fuzzy' phenotype. A fruit fly that is homozygous recessive for Gene F is covered in a dense layer of hairs, giving it a soft appearance. |
| Phenotype | Genotypes | Progeny Count | |
|---|---|---|---|
| bumpy, fuzzy | b | f | 8,555 |
| bumpy | b | + | 26,571 |
| fuzzy | + | f | 26,615 |
| wildtype | + | + | 8,259 |
| TOTAL = | 70,000 | ||
The resulting phenotypes are summarized in the table above.
With the progeny data from the table, calculate the genetic distance between the two genes, expressing your answer in centimorgans (cM)
8,259/70,000 = 8,259/70,000 = 0.1180 = 11.80 cM Incorrect 8,555/70,000 = 8,555/70,000 = 0.1222 = 12.22 cM Incorrect (8,259 + 8,555)/70,000 = 16,814/70,000 = 0.2402 = 24.02 cM Correct 26,571/70,000 = 26,571/70,000 = 0.3796 = 37.96 cM Incorrect 26,615/70,000 = 26,615/70,000 = 0.3802 = 38.02 cM Incorrect (8,555 + 26,571)/70,000 = 35,126/70,000 = 0.5018 = 50.18 cM Incorrect MCef1c_b35e
A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.
| Gene J is associated with the 'jerky' phenotype. A fruit fly that is homozygous recessive for Gene J moves in rapid and sudden movements, displaying an unpredictable flight pattern. | |
| Gene P is correlated with the 'prickly' phenotype. A fruit fly that is homozygous recessive for Gene P is covered with sharp bristles, giving it a spiky texture. |
| Phenotype | Genotypes | Progeny Count | |
|---|---|---|---|
| jerky, prickly | j | p | 10,375 |
| jerky | j | + | 3,943 |
| prickly | + | p | 3,974 |
| wildtype | + | + | 10,708 |
| TOTAL = | 29,000 | ||
The resulting phenotypes are summarized in the table above.
With the progeny data from the table, calculate the genetic distance between the two genes, expressing your answer in centimorgans (cM)
3,943/29,000 = 3,943/29,000 = 0.1360 = 13.60 cM Incorrect 3,974/29,000 = 3,974/29,000 = 0.1370 = 13.70 cM Incorrect (3,943 + 3,974)/29,000 = 7,917/29,000 = 0.2730 = 27.30 cM Correct 10,375/29,000 = 10,375/29,000 = 0.3578 = 35.78 cM Incorrect 10,708/29,000 = 10,708/29,000 = 0.3692 = 36.92 cM Incorrect (3,943 + 10,375)/29,000 = 14,318/29,000 = 0.4937 = 49.37 cM Incorrect MC57e7_29f6
A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.
| Gene F is connected with the 'fuzzy' phenotype. A fruit fly that is homozygous recessive for Gene F is covered in a dense layer of hairs, giving it a soft appearance. | |
| Gene N is affiliated with the 'nerdy' phenotype. A fruit fly that is homozygous recessive for Gene N has large, prominent eyes that stand out, much like thick-rimmed glasses. |
| Phenotype | Genotypes | Progeny Count | |
|---|---|---|---|
| fuzzy, nerdy | f | n | 114,602 |
| fuzzy | f | + | 25,254 |
| nerdy | + | n | 25,062 |
| wildtype | + | + | 115,082 |
| TOTAL = | 280,000 | ||
The resulting phenotypes are summarized in the table above.
With the progeny data from the table, calculate the genetic distance between the two genes, expressing your answer in centimorgans (cM)
25,062/280,000 = 25,062/280,000 = 0.0895 = 8.95 cM Incorrect 25,254/280,000 = 25,254/280,000 = 0.0902 = 9.02 cM Incorrect (25,062 + 25,254)/280,000 = 50,316/280,000 = 0.1797 = 17.97 cM Correct 114,602/280,000 = 114,602/280,000 = 0.4093 = 40.93 cM Incorrect 115,082/280,000 = 115,082/280,000 = 0.4110 = 41.10 cM Incorrect (25,062 + 114,602)/280,000 = 139,664/280,000 = 0.4988 = 49.88 cM Incorrect MC417f_73fa
A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.
| Gene B is connected with the 'bumpy' phenotype. A fruit fly that is homozygous recessive for Gene B has a skin texture that is not smooth, but rough with small bumps all over. | |
| Gene D is correlated with the 'dewy' phenotype. A fruit fly that is homozygous recessive for Gene D appears moist, with its body covered in tiny droplets of water. |
| Phenotype | Genotypes | Progeny Count | |
|---|---|---|---|
| bumpy, dewy | b | d | 13,668 |
| bumpy | b | + | 81,593 |
| dewy | + | d | 80,990 |
| wildtype | + | + | 13,749 |
| TOTAL = | 190,000 | ||
The resulting phenotypes are summarized in the table above.
With the progeny data from the table, calculate the genetic distance between the two genes, expressing your answer in centimorgans (cM)
13,668/190,000 = 13,668/190,000 = 0.0719 = 7.19 cM Incorrect 13,749/190,000 = 13,749/190,000 = 0.0724 = 7.24 cM Incorrect (13,668 + 13,749)/190,000 = 27,417/190,000 = 0.1443 = 14.43 cM Correct 80,990/190,000 = 80,990/190,000 = 0.4263 = 42.63 cM Incorrect 81,593/190,000 = 81,593/190,000 = 0.4294 = 42.94 cM Incorrect (13,749 + 80,990)/190,000 = 94,739/190,000 = 0.4986 = 49.86 cM Incorrect MC42e1_fc1f
A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.
| Gene N is connected with the 'nerdy' phenotype. A fruit fly that is homozygous recessive for Gene N has large, prominent eyes that stand out, much like thick-rimmed glasses. | |
| Gene R is associated with the 'rusty' phenotype. A fruit fly that is homozygous recessive for Gene R has a reddish-brown color, much like rusted iron metal. |
| Phenotype | Genotypes | Progeny Count | |
|---|---|---|---|
| nerdy, rusty | n | r | 10,089 |
| nerdy | n | + | 29,751 |
| rusty | + | r | 29,921 |
| wildtype | + | + | 10,239 |
| TOTAL = | 80,000 | ||
The resulting phenotypes are summarized in the table above.
With the progeny data from the table, calculate the genetic distance between the two genes, expressing your answer in centimorgans (cM)
10,089/80,000 = 10,089/80,000 = 0.1261 = 12.61 cM Incorrect 10,239/80,000 = 10,239/80,000 = 0.1280 = 12.80 cM Incorrect (10,089 + 10,239)/80,000 = 20,328/80,000 = 0.2541 = 25.41 cM Correct 29,751/80,000 = 29,751/80,000 = 0.3719 = 37.19 cM Incorrect 29,921/80,000 = 29,921/80,000 = 0.3740 = 37.40 cM Incorrect (10,089 + 29,751)/80,000 = 39,840/80,000 = 0.4980 = 49.80 cM Incorrect MC9a5c_c6b6
A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.
| Gene E is analogous to the 'eery' phenotype. A fruit fly that is homozygous recessive for Gene E appears to have something off, crooked limbs and other twisted appendages. | |
| Gene P is associated with the 'prickly' phenotype. A fruit fly that is homozygous recessive for Gene P is covered with sharp bristles, giving it a spiky texture. |
| Phenotype | Genotypes | Progeny Count | |
|---|---|---|---|
| eery, prickly | e | p | 7,519 |
| eery | e | + | 24,764 |
| prickly | + | p | 25,182 |
| wildtype | + | + | 7,535 |
| TOTAL = | 65,000 | ||
The resulting phenotypes are summarized in the table above.
With the progeny data from the table, calculate the genetic distance between the two genes, expressing your answer in centimorgans (cM)
7,519/65,000 = 7,519/65,000 = 0.1157 = 11.57 cM Incorrect 7,535/65,000 = 7,535/65,000 = 0.1159 = 11.59 cM Incorrect (7,519 + 7,535)/65,000 = 15,054/65,000 = 0.2316 = 23.16 cM Correct 24,764/65,000 = 24,764/65,000 = 0.3810 = 38.10 cM Incorrect 25,182/65,000 = 25,182/65,000 = 0.3874 = 38.74 cM Incorrect (7,535 + 24,764)/65,000 = 32,299/65,000 = 0.4969 = 49.69 cM Incorrect MC2425_5fdb
A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.
| Gene N is associated with the 'nerdy' phenotype. A fruit fly that is homozygous recessive for Gene N has large, prominent eyes that stand out, much like thick-rimmed glasses. | |
| Gene X is correlated with the 'xanthic' phenotype. A fruit fly that is homozygous recessive for Gene X has a fluorescent bright yellow coloring. |
| Phenotype | Genotypes | Progeny Count | |
|---|---|---|---|
| nerdy, xanthic | n | x | 31,233 |
| nerdy | n | + | 4,886 |
| xanthic | + | x | 4,800 |
| wildtype | + | + | 31,581 |
| TOTAL = | 72,500 | ||
The resulting phenotypes are summarized in the table above.
With the progeny data from the table, calculate the genetic distance between the two genes, expressing your answer in centimorgans (cM)
4,800/72,500 = 4,800/72,500 = 0.0662 = 6.62 cM Incorrect 4,886/72,500 = 4,886/72,500 = 0.0674 = 6.74 cM Incorrect (4,800 + 4,886)/72,500 = 9,686/72,500 = 0.1336 = 13.36 cM Correct 31,233/72,500 = 31,233/72,500 = 0.4308 = 43.08 cM Incorrect 31,581/72,500 = 31,581/72,500 = 0.4356 = 43.56 cM Incorrect (4,800 + 31,581)/72,500 = 36,381/72,500 = 0.5018 = 50.18 cM Incorrect MCcb31_c947
A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.
| Gene F is connected with the 'fuzzy' phenotype. A fruit fly that is homozygous recessive for Gene F is covered in a dense layer of hairs, giving it a soft appearance. | |
| Gene H is affiliated with the 'horsey' phenotype. A fruit fly that is homozygous recessive for Gene H is quite big and strong-looking, much larger than your typical fruit fly. |
| Phenotype | Genotypes | Progeny Count | |
|---|---|---|---|
| fuzzy, horsey | f | h | 1,426 |
| fuzzy | f | + | 4,848 |
| horsey | + | h | 4,847 |
| wildtype | + | + | 1,379 |
| TOTAL = | 12,500 | ||
The resulting phenotypes are summarized in the table above.
With the progeny data from the table, calculate the genetic distance between the two genes, expressing your answer in centimorgans (cM)
1,379/12,500 = 1,379/12,500 = 0.1103 = 11.03 cM Incorrect 1,426/12,500 = 1,426/12,500 = 0.1141 = 11.41 cM Incorrect (1,379 + 1,426)/12,500 = 2,805/12,500 = 0.2244 = 22.44 cM Correct 4,847/12,500 = 4,847/12,500 = 0.3878 = 38.78 cM Incorrect (1,379 + 4,847)/12,500 = 6,226/12,500 = 0.4981 = 49.81 cM Incorrect (1,426 + 4,848)/12,500 = 6,274/12,500 = 0.5019 = 50.19 cM Incorrect MCda26_bae5
A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.
| Gene F is associated with the 'fuzzy' phenotype. A fruit fly that is homozygous recessive for Gene F is covered in a dense layer of hairs, giving it a soft appearance. | |
| Gene J is correlated with the 'jerky' phenotype. A fruit fly that is homozygous recessive for Gene J moves in rapid and sudden movements, displaying an unpredictable flight pattern. |
| Phenotype | Genotypes | Progeny Count | |
|---|---|---|---|
| fuzzy, jerky | f | j | 58 |
| fuzzy | f | + | 220 |
| jerky | + | j | 251 |
| wildtype | + | + | 71 |
| TOTAL = | 600 | ||
The resulting phenotypes are summarized in the table above.
With the progeny data from the table, calculate the genetic distance between the two genes, expressing your answer in centimorgans (cM)
58/600 = 58/600 = 0.0967 = 9.67 cM Incorrect 71/600 = 71/600 = 0.1183 = 11.83 cM Incorrect (58 + 71)/600 = 129/600 = 0.2150 = 21.50 cM Correct 220/600 = 220/600 = 0.3667 = 36.67 cM Incorrect 251/600 = 251/600 = 0.4183 = 41.83 cM Incorrect (71 + 251)/600 = 322/600 = 0.5367 = 53.67 cM Incorrect MCf5f3_966e
A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.
| Gene K is associated with the 'kidney' phenotype. A fruit fly that is homozygous recessive for Gene K has a body shape that is curved, similar to a kidney bean. | |
| Gene P is affiliated with the 'prickly' phenotype. A fruit fly that is homozygous recessive for Gene P is covered with sharp bristles, giving it a spiky texture. |
| Phenotype | Genotypes | Progeny Count | |
|---|---|---|---|
| kidney, prickly | k | p | 241 |
| kidney | k | + | 872 |
| prickly | + | p | 866 |
| wildtype | + | + | 221 |
| TOTAL = | 2,200 | ||
The resulting phenotypes are summarized in the table above.
With the progeny data from the table, calculate the genetic distance between the two genes, expressing your answer in centimorgans (cM)
221/2,200 = 221/2,200 = 0.1005 = 10.05 cM Incorrect 241/2,200 = 241/2,200 = 0.1095 = 10.95 cM Incorrect (221 + 241)/2,200 = 462/2,200 = 0.2100 = 21.00 cM Correct 866/2,200 = 866/2,200 = 0.3936 = 39.36 cM Incorrect 872/2,200 = 872/2,200 = 0.3964 = 39.64 cM Incorrect (241 + 872)/2,200 = 1,113/2,200 = 0.5059 = 50.59 cM Incorrect MCbc83_6b8e
A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.
| Gene T is linked with the 'tipsy' phenotype. A fruit fly that is homozygous recessive for Gene T moves in an erratic path, suggesting a lack of coordination, as if intoxicated. | |
| Gene W is analogous to the 'waxy' phenotype. A fruit fly that is homozygous recessive for Gene W has a thick protective layer that is water resistant and opague. |
| Phenotype | Genotypes | Progeny Count | |
|---|---|---|---|
| tipsy, waxy | t | w | 30,873 |
| tipsy | t | + | 83,969 |
| waxy | + | w | 83,632 |
| wildtype | + | + | 31,526 |
| TOTAL = | 230,000 | ||
The resulting phenotypes are summarized in the table above.
With the progeny data from the table, calculate the genetic distance between the two genes, expressing your answer in centimorgans (cM)
30,873/230,000 = 30,873/230,000 = 0.1342 = 13.42 cM Incorrect 31,526/230,000 = 31,526/230,000 = 0.1371 = 13.71 cM Incorrect (30,873 + 31,526)/230,000 = 62,399/230,000 = 0.2713 = 27.13 cM Correct 83,632/230,000 = 83,632/230,000 = 0.3636 = 36.36 cM Incorrect 83,969/230,000 = 83,969/230,000 = 0.3651 = 36.51 cM Incorrect (30,873 + 83,969)/230,000 = 114,842/230,000 = 0.4993 = 49.93 cM Incorrect MCdd48_c9cb
A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.
| Gene E is analogous to the 'eery' phenotype. A fruit fly that is homozygous recessive for Gene E appears to have something off, crooked limbs and other twisted appendages. | |
| Gene T is related to the 'tipsy' phenotype. A fruit fly that is homozygous recessive for Gene T moves in an erratic path, suggesting a lack of coordination, as if intoxicated. |
| Phenotype | Genotypes | Progeny Count | |
|---|---|---|---|
| eery, tipsy | e | t | 24,225 |
| eery | e | + | 3,240 |
| tipsy | + | t | 3,338 |
| wildtype | + | + | 24,197 |
| TOTAL = | 55,000 | ||
The resulting phenotypes are summarized in the table above.
With the progeny data from the table, calculate the genetic distance between the two genes, expressing your answer in centimorgans (cM)
3,240/55,000 = 3,240/55,000 = 0.0589 = 5.89 cM Incorrect 3,338/55,000 = 3,338/55,000 = 0.0607 = 6.07 cM Incorrect (3,240 + 3,338)/55,000 = 6,578/55,000 = 0.1196 = 11.96 cM Correct 24,197/55,000 = 24,197/55,000 = 0.4399 = 43.99 cM Incorrect 24,225/55,000 = 24,225/55,000 = 0.4405 = 44.05 cM Incorrect (3,338 + 24,197)/55,000 = 27,535/55,000 = 0.5006 = 50.06 cM Incorrect MC68f8_71d8
A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.
| Gene R is related to the 'rusty' phenotype. A fruit fly that is homozygous recessive for Gene R has a reddish-brown color, much like rusted iron metal. | |
| Gene W is related to the 'waxy' phenotype. A fruit fly that is homozygous recessive for Gene W has a thick protective layer that is water resistant and opague. |
| Phenotype | Genotypes | Progeny Count | |
|---|---|---|---|
| rusty, waxy | r | w | 5,223 |
| rusty | r | + | 24,621 |
| waxy | + | w | 24,795 |
| wildtype | + | + | 5,361 |
| TOTAL = | 60,000 | ||
The resulting phenotypes are summarized in the table above.
With the progeny data from the table, calculate the genetic distance between the two genes, expressing your answer in centimorgans (cM)
5,223/60,000 = 5,223/60,000 = 0.0871 = 8.71 cM Incorrect 5,361/60,000 = 5,361/60,000 = 0.0893 = 8.94 cM Incorrect (5,223 + 5,361)/60,000 = 10,584/60,000 = 0.1764 = 17.64 cM Correct 24,621/60,000 = 24,621/60,000 = 0.4103 = 41.03 cM Incorrect 24,795/60,000 = 24,795/60,000 = 0.4133 = 41.33 cM Incorrect (5,223 + 24,621)/60,000 = 29,844/60,000 = 0.4974 = 49.74 cM Incorrect MC53f1_87df
A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.
| Gene D is affiliated with the 'dewy' phenotype. A fruit fly that is homozygous recessive for Gene D appears moist, with its body covered in tiny droplets of water. | |
| Gene Y is related to the 'yucky' phenotype. A fruit fly that is homozygous recessive for Gene Y gives off an unpleasant odor and has a generally unappealing look. |
| Phenotype | Genotypes | Progeny Count | |
|---|---|---|---|
| dewy, yucky | d | y | 23,593 |
| dewy | d | + | 6,450 |
| yucky | + | y | 6,426 |
| wildtype | + | + | 23,531 |
| TOTAL = | 60,000 | ||
The resulting phenotypes are summarized in the table above.
With the progeny data from the table, calculate the genetic distance between the two genes, expressing your answer in centimorgans (cM)
6,426/60,000 = 6,426/60,000 = 0.1071 = 10.71 cM Incorrect 6,450/60,000 = 6,450/60,000 = 0.1075 = 10.75 cM Incorrect (6,426 + 6,450)/60,000 = 12,876/60,000 = 0.2146 = 21.46 cM Correct 23,531/60,000 = 23,531/60,000 = 0.3922 = 39.22 cM Incorrect 23,593/60,000 = 23,593/60,000 = 0.3932 = 39.32 cM Incorrect (6,450 + 23,593)/60,000 = 30,043/60,000 = 0.5007 = 50.07 cM Incorrect MC926b_c190
A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.
| Gene C is affiliated with the 'chummy' phenotype. A fruit fly that is homozygous recessive for Gene C shows behavior where it always maintains a close distance to other flies. | |
| Gene N is affiliated with the 'nerdy' phenotype. A fruit fly that is homozygous recessive for Gene N has large, prominent eyes that stand out, much like thick-rimmed glasses. |
| Phenotype | Genotypes | Progeny Count | |
|---|---|---|---|
| chummy, nerdy | c | n | 16,251 |
| chummy | c | + | 8,743 |
| nerdy | + | n | 8,627 |
| wildtype | + | + | 16,379 |
| TOTAL = | 50,000 | ||
The resulting phenotypes are summarized in the table above.
With the progeny data from the table, calculate the genetic distance between the two genes, expressing your answer in centimorgans (cM)
8,627/50,000 = 8,627/50,000 = 0.1725 = 17.25 cM Incorrect 8,743/50,000 = 8,743/50,000 = 0.1749 = 17.49 cM Incorrect 16,251/50,000 = 16,251/50,000 = 0.3250 = 32.50 cM Incorrect 16,379/50,000 = 16,379/50,000 = 0.3276 = 32.76 cM Incorrect (8,627 + 8,743)/50,000 = 17,370/50,000 = 0.3474 = 34.74 cM Correct (8,627 + 16,251)/50,000 = 24,878/50,000 = 0.4976 = 49.76 cM Incorrect MC58ff_95b0
A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.
| Gene M is affiliated with the 'mushy' phenotype. A fruit fly that is homozygous recessive for Gene M feels soft to the touch and unusually squishy, unlike the usual firmness. | |
| Gene R is linked with the 'rusty' phenotype. A fruit fly that is homozygous recessive for Gene R has a reddish-brown color, much like rusted iron metal. |
| Phenotype | Genotypes | Progeny Count | |
|---|---|---|---|
| mushy, rusty | m | r | 1,016 |
| mushy | m | + | 8,459 |
| rusty | + | r | 8,470 |
| wildtype | + | + | 1,055 |
| TOTAL = | 19,000 | ||
The resulting phenotypes are summarized in the table above.
With the progeny data from the table, calculate the genetic distance between the two genes, expressing your answer in centimorgans (cM)
1,016/19,000 = 1,016/19,000 = 0.0535 = 5.35 cM Incorrect 1,055/19,000 = 1,055/19,000 = 0.0555 = 5.55 cM Incorrect (1,016 + 1,055)/19,000 = 2,071/19,000 = 0.1090 = 10.90 cM Correct 8,459/19,000 = 8,459/19,000 = 0.4452 = 44.52 cM Incorrect 8,470/19,000 = 8,470/19,000 = 0.4458 = 44.58 cM Incorrect (1,016 + 8,459)/19,000 = 9,475/19,000 = 0.4987 = 49.87 cM Incorrect MC749b_950f
A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.
| Gene E is analogous to the 'eery' phenotype. A fruit fly that is homozygous recessive for Gene E appears to have something off, crooked limbs and other twisted appendages. | |
| Gene T is analogous to the 'tipsy' phenotype. A fruit fly that is homozygous recessive for Gene T moves in an erratic path, suggesting a lack of coordination, as if intoxicated. |
| Phenotype | Genotypes | Progeny Count | |
|---|---|---|---|
| eery, tipsy | e | t | 38,855 |
| eery | e | + | 6,063 |
| tipsy | + | t | 5,934 |
| wildtype | + | + | 39,148 |
| TOTAL = | 90,000 | ||
The resulting phenotypes are summarized in the table above.
With the progeny data from the table, calculate the genetic distance between the two genes, expressing your answer in centimorgans (cM)
5,934/90,000 = 5,934/90,000 = 0.0659 = 6.59 cM Incorrect 6,063/90,000 = 6,063/90,000 = 0.0674 = 6.74 cM Incorrect (5,934 + 6,063)/90,000 = 11,997/90,000 = 0.1333 = 13.33 cM Correct 38,855/90,000 = 38,855/90,000 = 0.4317 = 43.17 cM Incorrect 39,148/90,000 = 39,148/90,000 = 0.4350 = 43.50 cM Incorrect (6,063 + 39,148)/90,000 = 45,211/90,000 = 0.5023 = 50.23 cM Incorrect MCe2fd_9789
A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.
| Gene E is analogous to the 'eery' phenotype. A fruit fly that is homozygous recessive for Gene E appears to have something off, crooked limbs and other twisted appendages. | |
| Gene T is linked with the 'tipsy' phenotype. A fruit fly that is homozygous recessive for Gene T moves in an erratic path, suggesting a lack of coordination, as if intoxicated. |
| Phenotype | Genotypes | Progeny Count | |
|---|---|---|---|
| eery, tipsy | e | t | 2,300 |
| eery | e | + | 910 |
| tipsy | + | t | 858 |
| wildtype | + | + | 2,432 |
| TOTAL = | 6,500 | ||
The resulting phenotypes are summarized in the table above.
With the progeny data from the table, calculate the genetic distance between the two genes, expressing your answer in centimorgans (cM)
858/6,500 = 858/6,500 = 0.1320 = 13.20 cM Incorrect 910/6,500 = 910/6,500 = 0.1400 = 14.00 cM Incorrect (858 + 910)/6,500 = 1,768/6,500 = 0.2720 = 27.20 cM Correct 2,300/6,500 = 2,300/6,500 = 0.3538 = 35.38 cM Incorrect 2,432/6,500 = 2,432/6,500 = 0.3742 = 37.42 cM Incorrect (858 + 2,432)/6,500 = 3,290/6,500 = 0.5062 = 50.62 cM Incorrect MC4a05_28a5
A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.
| Gene B is correlated with the 'bumpy' phenotype. A fruit fly that is homozygous recessive for Gene B has a skin texture that is not smooth, but rough with small bumps all over. | |
| Gene K is correlated with the 'kidney' phenotype. A fruit fly that is homozygous recessive for Gene K has a body shape that is curved, similar to a kidney bean. |
| Phenotype | Genotypes | Progeny Count | |
|---|---|---|---|
| bumpy, kidney | b | k | 687 |
| bumpy | b | + | 91 |
| kidney | + | k | 117 |
| wildtype | + | + | 705 |
| TOTAL = | 1,600 | ||
The resulting phenotypes are summarized in the table above.
With the progeny data from the table, calculate the genetic distance between the two genes, expressing your answer in centimorgans (cM)
91/1,600 = 91/1,600 = 0.0569 = 5.69 cM Incorrect 117/1,600 = 117/1,600 = 0.0731 = 7.31 cM Incorrect (91 + 117)/1,600 = 208/1,600 = 0.1300 = 13.00 cM Correct 687/1,600 = 687/1,600 = 0.4294 = 42.94 cM Incorrect 705/1,600 = 705/1,600 = 0.4406 = 44.06 cM Incorrect (91 + 687)/1,600 = 778/1,600 = 0.4863 = 48.62 cM Incorrect MC5559_c77a
A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.
| Gene H is correlated with the 'horsey' phenotype. A fruit fly that is homozygous recessive for Gene H is quite big and strong-looking, much larger than your typical fruit fly. | |
| Gene Y is linked with the 'yucky' phenotype. A fruit fly that is homozygous recessive for Gene Y gives off an unpleasant odor and has a generally unappealing look. |
| Phenotype | Genotypes | Progeny Count | |
|---|---|---|---|
| horsey, yucky | h | y | 2,931 |
| horsey | h | + | 11,992 |
| yucky | + | y | 12,101 |
| wildtype | + | + | 2,976 |
| TOTAL = | 30,000 | ||
The resulting phenotypes are summarized in the table above.
With the progeny data from the table, calculate the genetic distance between the two genes, expressing your answer in centimorgans (cM)
2,931/30,000 = 2,931/30,000 = 0.0977 = 9.77 cM Incorrect 2,976/30,000 = 2,976/30,000 = 0.0992 = 9.92 cM Incorrect (2,931 + 2,976)/30,000 = 5,907/30,000 = 0.1969 = 19.69 cM Correct 11,992/30,000 = 11,992/30,000 = 0.3997 = 39.97 cM Incorrect 12,101/30,000 = 12,101/30,000 = 0.4034 = 40.34 cM Incorrect (2,931 + 11,992)/30,000 = 14,923/30,000 = 0.4974 = 49.74 cM Incorrect MC36d4_7913
A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.
| Gene T is connected with the 'tipsy' phenotype. A fruit fly that is homozygous recessive for Gene T moves in an erratic path, suggesting a lack of coordination, as if intoxicated. | |
| Gene X is analogous to the 'xanthic' phenotype. A fruit fly that is homozygous recessive for Gene X has a fluorescent bright yellow coloring. |
| Phenotype | Genotypes | Progeny Count | |
|---|---|---|---|
| tipsy, xanthic | t | x | 9,139 |
| tipsy | t | + | 35,864 |
| xanthic | + | x | 35,893 |
| wildtype | + | + | 9,104 |
| TOTAL = | 90,000 | ||
The resulting phenotypes are summarized in the table above.
With the progeny data from the table, calculate the genetic distance between the two genes, expressing your answer in centimorgans (cM)
9,104/90,000 = 9,104/90,000 = 0.1012 = 10.12 cM Incorrect 9,139/90,000 = 9,139/90,000 = 0.1015 = 10.15 cM Incorrect (9,104 + 9,139)/90,000 = 18,243/90,000 = 0.2027 = 20.27 cM Correct 35,864/90,000 = 35,864/90,000 = 0.3985 = 39.85 cM Incorrect 35,893/90,000 = 35,893/90,000 = 0.3988 = 39.88 cM Incorrect (9,104 + 35,893)/90,000 = 44,997/90,000 = 0.5000 = 50.00 cM Incorrect MC1585_8ea4
A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.
| Gene F is related to the 'fuzzy' phenotype. A fruit fly that is homozygous recessive for Gene F is covered in a dense layer of hairs, giving it a soft appearance. | |
| Gene K is analogous to the 'kidney' phenotype. A fruit fly that is homozygous recessive for Gene K has a body shape that is curved, similar to a kidney bean. |
| Phenotype | Genotypes | Progeny Count | |
|---|---|---|---|
| fuzzy, kidney | f | k | 30,208 |
| fuzzy | f | + | 5,848 |
| kidney | + | k | 5,984 |
| wildtype | + | + | 30,460 |
| TOTAL = | 72,500 | ||
The resulting phenotypes are summarized in the table above.
With the progeny data from the table, calculate the genetic distance between the two genes, expressing your answer in centimorgans (cM)
5,848/72,500 = 5,848/72,500 = 0.0807 = 8.07 cM Incorrect 5,984/72,500 = 5,984/72,500 = 0.0825 = 8.25 cM Incorrect (5,848 + 5,984)/72,500 = 11,832/72,500 = 0.1632 = 16.32 cM Correct 30,208/72,500 = 30,208/72,500 = 0.4167 = 41.67 cM Incorrect 30,460/72,500 = 30,460/72,500 = 0.4201 = 42.01 cM Incorrect (5,848 + 30,460)/72,500 = 36,308/72,500 = 0.5008 = 50.08 cM Incorrect MCfd72_02b0
A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.
| Gene D is correlated with the 'dewy' phenotype. A fruit fly that is homozygous recessive for Gene D appears moist, with its body covered in tiny droplets of water. | |
| Gene H is correlated with the 'horsey' phenotype. A fruit fly that is homozygous recessive for Gene H is quite big and strong-looking, much larger than your typical fruit fly. |
| Phenotype | Genotypes | Progeny Count | |
|---|---|---|---|
| dewy, horsey | d | h | 12,916 |
| dewy | d | + | 37,107 |
| horsey | + | h | 36,963 |
| wildtype | + | + | 13,014 |
| TOTAL = | 100,000 | ||
The resulting phenotypes are summarized in the table above.
With the progeny data from the table, calculate the genetic distance between the two genes, expressing your answer in centimorgans (cM)
12,916/100,000 = 12,916/100,000 = 0.1292 = 12.92 cM Incorrect 13,014/100,000 = 13,014/100,000 = 0.1301 = 13.01 cM Incorrect (12,916 + 13,014)/100,000 = 25,930/100,000 = 0.2593 = 25.93 cM Correct 36,963/100,000 = 36,963/100,000 = 0.3696 = 36.96 cM Incorrect 37,107/100,000 = 37,107/100,000 = 0.3711 = 37.11 cM Incorrect (13,014 + 36,963)/100,000 = 49,977/100,000 = 0.4998 = 49.98 cM Incorrect MCf010_1bec
A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.
| Gene E is connected with the 'eery' phenotype. A fruit fly that is homozygous recessive for Gene E appears to have something off, crooked limbs and other twisted appendages. | |
| Gene F is analogous to the 'fuzzy' phenotype. A fruit fly that is homozygous recessive for Gene F is covered in a dense layer of hairs, giving it a soft appearance. |
| Phenotype | Genotypes | Progeny Count | |
|---|---|---|---|
| eery, fuzzy | e | f | 16,978 |
| eery | e | + | 28,083 |
| fuzzy | + | f | 27,915 |
| wildtype | + | + | 17,024 |
| TOTAL = | 90,000 | ||
The resulting phenotypes are summarized in the table above.
With the progeny data from the table, calculate the genetic distance between the two genes, expressing your answer in centimorgans (cM)
16,978/90,000 = 16,978/90,000 = 0.1886 = 18.86 cM Incorrect 17,024/90,000 = 17,024/90,000 = 0.1892 = 18.92 cM Incorrect 27,915/90,000 = 27,915/90,000 = 0.3102 = 31.02 cM Incorrect 28,083/90,000 = 28,083/90,000 = 0.3120 = 31.20 cM Incorrect (16,978 + 17,024)/90,000 = 34,002/90,000 = 0.3778 = 37.78 cM Correct (16,978 + 27,915)/90,000 = 44,893/90,000 = 0.4988 = 49.88 cM Incorrect MC268e_d7d0
A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.
| Gene C is affiliated with the 'chummy' phenotype. A fruit fly that is homozygous recessive for Gene C shows behavior where it always maintains a close distance to other flies. | |
| Gene E is correlated with the 'eery' phenotype. A fruit fly that is homozygous recessive for Gene E appears to have something off, crooked limbs and other twisted appendages. |
| Phenotype | Genotypes | Progeny Count | |
|---|---|---|---|
| chummy, eery | c | e | 21,931 |
| chummy | c | + | 7,977 |
| eery | + | e | 8,127 |
| wildtype | + | + | 21,965 |
| TOTAL = | 60,000 | ||
The resulting phenotypes are summarized in the table above.
With the progeny data from the table, calculate the genetic distance between the two genes, expressing your answer in centimorgans (cM)
7,977/60,000 = 7,977/60,000 = 0.1330 = 13.30 cM Incorrect 8,127/60,000 = 8,127/60,000 = 0.1354 = 13.54 cM Incorrect (7,977 + 8,127)/60,000 = 16,104/60,000 = 0.2684 = 26.84 cM Correct 21,931/60,000 = 21,931/60,000 = 0.3655 = 36.55 cM Incorrect 21,965/60,000 = 21,965/60,000 = 0.3661 = 36.61 cM Incorrect (8,127 + 21,931)/60,000 = 30,058/60,000 = 0.5010 = 50.10 cM Incorrect MCfa59_6871
A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.
| Gene H is affiliated with the 'horsey' phenotype. A fruit fly that is homozygous recessive for Gene H is quite big and strong-looking, much larger than your typical fruit fly. | |
| Gene Y is associated with the 'yucky' phenotype. A fruit fly that is homozygous recessive for Gene Y gives off an unpleasant odor and has a generally unappealing look. |
| Phenotype | Genotypes | Progeny Count | |
|---|---|---|---|
| horsey, yucky | h | y | 74,866 |
| horsey | h | + | 25,072 |
| yucky | + | y | 25,068 |
| wildtype | + | + | 74,994 |
| TOTAL = | 200,000 | ||
The resulting phenotypes are summarized in the table above.
With the progeny data from the table, calculate the genetic distance between the two genes, expressing your answer in centimorgans (cM)
25,068/200,000 = 25,068/200,000 = 0.1253 = 12.53 cM Incorrect 25,072/200,000 = 25,072/200,000 = 0.1254 = 12.54 cM Incorrect (25,068 + 25,072)/200,000 = 50,140/200,000 = 0.2507 = 25.07 cM Correct 74,866/200,000 = 74,866/200,000 = 0.3743 = 37.43 cM Incorrect 74,994/200,000 = 74,994/200,000 = 0.3750 = 37.50 cM Incorrect (25,072 + 74,866)/200,000 = 99,938/200,000 = 0.4997 = 49.97 cM Incorrect MC268b_17c4
A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.
| Gene B is analogous to the 'bumpy' phenotype. A fruit fly that is homozygous recessive for Gene B has a skin texture that is not smooth, but rough with small bumps all over. | |
| Gene X is affiliated with the 'xanthic' phenotype. A fruit fly that is homozygous recessive for Gene X has a fluorescent bright yellow coloring. |
| Phenotype | Genotypes | Progeny Count | |
|---|---|---|---|
| bumpy, xanthic | b | x | 8,720 |
| bumpy | b | + | 1,390 |
| xanthic | + | x | 1,340 |
| wildtype | + | + | 8,550 |
| TOTAL = | 20,000 | ||
The resulting phenotypes are summarized in the table above.
With the progeny data from the table, calculate the genetic distance between the two genes, expressing your answer in centimorgans (cM)
1,340/20,000 = 1,340/20,000 = 0.0670 = 6.70 cM Incorrect 1,390/20,000 = 1,390/20,000 = 0.0695 = 6.95 cM Incorrect (1,340 + 1,390)/20,000 = 2,730/20,000 = 0.1365 = 13.65 cM Correct 8,550/20,000 = 8,550/20,000 = 0.4275 = 42.75 cM Incorrect 8,720/20,000 = 8,720/20,000 = 0.4360 = 43.60 cM Incorrect (1,390 + 8,720)/20,000 = 10,110/20,000 = 0.5055 = 50.55 cM Incorrect MC00fe_003b
A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.
| Gene H is connected with the 'horsey' phenotype. A fruit fly that is homozygous recessive for Gene H is quite big and strong-looking, much larger than your typical fruit fly. | |
| Gene M is affiliated with the 'mushy' phenotype. A fruit fly that is homozygous recessive for Gene M feels soft to the touch and unusually squishy, unlike the usual firmness. |
| Phenotype | Genotypes | Progeny Count | |
|---|---|---|---|
| horsey, mushy | h | m | 2,755 |
| horsey | h | + | 17,218 |
| mushy | + | m | 17,294 |
| wildtype | + | + | 2,733 |
| TOTAL = | 40,000 | ||
The resulting phenotypes are summarized in the table above.
With the progeny data from the table, calculate the genetic distance between the two genes, expressing your answer in centimorgans (cM)
2,733/40,000 = 2,733/40,000 = 0.0683 = 6.83 cM Incorrect 2,755/40,000 = 2,755/40,000 = 0.0689 = 6.89 cM Incorrect (2,733 + 2,755)/40,000 = 5,488/40,000 = 0.1372 = 13.72 cM Correct 17,218/40,000 = 17,218/40,000 = 0.4304 = 43.05 cM Incorrect 17,294/40,000 = 17,294/40,000 = 0.4324 = 43.23 cM Incorrect (2,733 + 17,294)/40,000 = 20,027/40,000 = 0.5007 = 50.07 cM Incorrect MC60c5_d49b
A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.
| Gene T is linked with the 'tipsy' phenotype. A fruit fly that is homozygous recessive for Gene T moves in an erratic path, suggesting a lack of coordination, as if intoxicated. | |
| Gene Y is analogous to the 'yucky' phenotype. A fruit fly that is homozygous recessive for Gene Y gives off an unpleasant odor and has a generally unappealing look. |
| Phenotype | Genotypes | Progeny Count | |
|---|---|---|---|
| tipsy, yucky | t | y | 51,664 |
| tipsy | t | + | 84,026 |
| yucky | + | y | 83,293 |
| wildtype | + | + | 51,017 |
| TOTAL = | 270,000 | ||
The resulting phenotypes are summarized in the table above.
With the progeny data from the table, calculate the genetic distance between the two genes, expressing your answer in centimorgans (cM)
51,017/270,000 = 51,017/270,000 = 0.1890 = 18.90 cM Incorrect 51,664/270,000 = 51,664/270,000 = 0.1913 = 19.13 cM Incorrect 83,293/270,000 = 83,293/270,000 = 0.3085 = 30.85 cM Incorrect 84,026/270,000 = 84,026/270,000 = 0.3112 = 31.12 cM Incorrect (51,017 + 51,664)/270,000 = 102,681/270,000 = 0.3803 = 38.03 cM Correct (51,664 + 83,293)/270,000 = 134,957/270,000 = 0.4998 = 49.98 cM Incorrect MCebda_aa0a
A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.
| Gene D is affiliated with the 'dewy' phenotype. A fruit fly that is homozygous recessive for Gene D appears moist, with its body covered in tiny droplets of water. | |
| Gene M is associated with the 'mushy' phenotype. A fruit fly that is homozygous recessive for Gene M feels soft to the touch and unusually squishy, unlike the usual firmness. |
| Phenotype | Genotypes | Progeny Count | |
|---|---|---|---|
| dewy, mushy | d | m | 18,825 |
| dewy | d | + | 8,477 |
| mushy | + | m | 8,236 |
| wildtype | + | + | 18,462 |
| TOTAL = | 54,000 | ||
The resulting phenotypes are summarized in the table above.
With the progeny data from the table, calculate the genetic distance between the two genes, expressing your answer in centimorgans (cM)
8,236/54,000 = 8,236/54,000 = 0.1525 = 15.25 cM Incorrect 8,477/54,000 = 8,477/54,000 = 0.1570 = 15.70 cM Incorrect (8,236 + 8,477)/54,000 = 16,713/54,000 = 0.3095 = 30.95 cM Correct 18,462/54,000 = 18,462/54,000 = 0.3419 = 34.19 cM Incorrect 18,825/54,000 = 18,825/54,000 = 0.3486 = 34.86 cM Incorrect (8,236 + 18,462)/54,000 = 26,698/54,000 = 0.4944 = 49.44 cM Incorrect MC371e_56a0
A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.
| Gene H is analogous to the 'horsey' phenotype. A fruit fly that is homozygous recessive for Gene H is quite big and strong-looking, much larger than your typical fruit fly. | |
| Gene P is connected with the 'prickly' phenotype. A fruit fly that is homozygous recessive for Gene P is covered with sharp bristles, giving it a spiky texture. |
| Phenotype | Genotypes | Progeny Count | |
|---|---|---|---|
| horsey, prickly | h | p | 9,630 |
| horsey | h | + | 95,565 |
| prickly | + | p | 95,388 |
| wildtype | + | + | 9,417 |
| TOTAL = | 210,000 | ||
The resulting phenotypes are summarized in the table above.
With the progeny data from the table, calculate the genetic distance between the two genes, expressing your answer in centimorgans (cM)
9,417/210,000 = 9,417/210,000 = 0.0448 = 4.48 cM Incorrect 9,630/210,000 = 9,630/210,000 = 0.0459 = 4.59 cM Incorrect (9,417 + 9,630)/210,000 = 19,047/210,000 = 0.0907 = 9.07 cM Correct 95,388/210,000 = 95,388/210,000 = 0.4542 = 45.42 cM Incorrect 95,565/210,000 = 95,565/210,000 = 0.4551 = 45.51 cM Incorrect (9,417 + 95,565)/210,000 = 104,982/210,000 = 0.4999 = 49.99 cM Incorrect